Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(){
  int N;cin>>N;double p[N],DP[N+1][N+1],sum=0;
  for(int i=0;i<N;i++)cin>>p[i];DP[0][0]=1;
  for(int i=1;i<=N;i++){
    DP[i][0]=DP[i-1][0]*(1-p[i-1]);DP[i][i]=DP[i-1][i-1]*p[i-1];
    for(int j=1;j<i;j++)DP[i][j]=DP[i-1][j-1]*p[i-1]+DP[i-1][j]*(1-p[i-1]);
  }
  for(int i=(N+1)/2;i<=N;i++)sum+=DP[N][i];
  printf("%.10lf\n",sum);
}